How is the Angular Velocity of a Planetary Gear Determined?

[rara]

Gear C has radius 0.12 m and rotates with angular velocity 1.9 k rad/s. The connecting link rotates at angular velocity 1.5 k rad/s. Gear D has radius 0.05 m. Find the angular velocity of gear D (in rad/s). Note that gear C is pinned to ground and gear D is a planetary gear.

Since gear c is pinned, the gear ratio ωc/ωd=rd/rc does not directly apply here right? I wonder if the correct solution would be ωd=ωc x rcd / rd. I got -2.125 rad/s. Thanks.

 

[TSny]

Hello.

At the point of contact of the gears, suppose you pick a point $c$ on the rim of gear C and a point $d$ on the rim of gear D as shown.

Let $\vec{V_c}$ and $\vec{V_d}$ be the velocities of those points, respectively, as measured relative to the fixed point at the center of gear C.

How are $\vec{V_c}$ and $\vec{V_d}$ related?

 

[rara]

Vc and Vd are the same

 

[TSny]

OK. So, $\vec{V_c}$ = $\vec{V_d}$. If you can express each of these velocities in terms of the angular velocities of the gears ($\omega_C$ and $\omega_D$) and the angular velocity of the link ($\omega_L$), then you would have an equation that you could solve for $\omega_D$.

Start with the velocity vector $\vec{V_c}$. Suppose you introduce a unit vector $\hat{t}$ that is tangent to the gears at the point of contact, as shown. How can you express $\vec{V_c}$ in terms of $\omega_C$, $r_C$, and $\hat{t}$?

 

[rezeew]

I can confirm that your approach to finding the angular velocity of gear D is correct. The gear ratio formula ωc/ωd=rd/rc does not apply in this case since gear C is pinned and gear D is a planetary gear. Your solution of ωd=ωc x rcd / rd is the correct way to calculate the angular velocity of gear D. Plugging in the given values, we get ωd=(1.9 k rad/s) x (0.12 m) / (0.05 m) = 4.56 k rad/s or approximately 2.125 rad/s. This angular velocity represents the rate at which gear D is rotating, taking into account its smaller radius and the rotational speed of the connecting link.